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Leto [7]
2 years ago
8

$24 saved after 3 weeks; $52 saved after 7 weeks. are these ratios equivalent

Mathematics
2 answers:
tankabanditka [31]2 years ago
8 0
Add 24+24=48 then 24/3= 8 add 48+8 = 56
SIZIF [17.4K]2 years ago
4 0
To figure out this question, I first divided 24 into 3, which gives me an answer of 8. then I divided 52 into 7 , which also is 8 . because the two numbers are the same, I know the ratios are equivalent.
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The blood platelet counts of a group of women have a​ bell-shaped distribution with a mean of 247.9 and a standard deviation of
labwork [276]

Answer:

A) Approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) approximate percentage of women with platelet counts between 53.8 and 442.0 = 99.7%

Step-by-step explanation:

We are given;

mean;μ = 247.9

standard deviation;σ = 64.7

A) We want to find the approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3.

Now, from the image attached, we can see that from the empirical curve, the probability of 1 standard deviation from the mean is (34% + 34%) = 68 %.

While probability of 2 standard deviations from the mean is (13.5% + 34% + 34% + 13.5%) = 95%

Thus, approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) Now, we want to find the approximate percentage of women with platelet counts between 53.8 and 442.0.

53.8 and 442.0 represents 3 standard deviations from the mean.

Let's confirm that.

Since mean;μ = 247.9

standard deviation;σ = 64.7 ;

μ = 247.9

σ = 64.7

μ + 3σ = 247.9 + 3(64.7) = 442

Also;

μ - 3σ = 247.9 - 3(64.7) = 53.8

Again from the empirical curve attached, we cans that at 3 standard deviations from the mean, we have a percentage probability of;

(2.35% + 13.5% + 34% + 34% + 13.5% + 2.35%) = 99.7%

5 0
1 year ago
Jenna eats dinner at a restaurant. Her bill is $14.50 Jenna pays her bill and leaves a tip. Write an inequality to represent the
Sophie [7]

Answer:

Step-by-step explanation:

x>14.50+y

7 0
2 years ago
Anika wants to determine The maximum number of tulip bulbs (T) she can purchase if each bulb costs $1.50. She will also need to
PtichkaEL [24]
T( 1.50+ 1.25) + 10.00 < 20.00 2.75t + 10.00 < 20.00 2.75t < 10.00 T < 3.63 Partial bulbs and puts can't be bought, so Anika cannot spent the full $20.00. Therefore, t < 3 if t can be a whole number.
6 0
2 years ago
Read 2 more answers
1/4 of the roses are red, 1/3 of the remainder are yellow and the rest are pink. There are 24 more pink roses than red roses. Ho
Alika [10]
<h3>There are 96 roses altogether</h3>

<em><u>Solution:</u></em>

Let "x" be the number of roses

From given,

<em><u>1/4 of the roses are red</u></em>

Red\ roses = \frac{1}{4} \times x\\\\Red\ roses = \frac{x}{4}

<em><u>1/3 of the remainder are yellow</u></em>

Remaining = x - \frac{x}{4}\\\\Remaining = \frac{3x}{4}

Therefore,

Yellow\ roses = \frac{1}{3} \times \frac{3x}{4}\\\\Yellow\ roses = \frac{x}{4}

<em><u>Rest are pink</u></em>

Remaining = \frac{3x}{4} - \frac{x}{4}\\\\Remaining = \frac{2x}{4}\\\\Remaining = \frac{x}{2}

Therefore,

Pink\ Roses = \frac{x}{2}

There are 24 more pink roses than red roses

Therefore,

Number of pink roses = 24 + red roses

\frac{x}{2} = 24 + \frac{x}{4}\\\\\frac{x}{2} -  \frac{x}{4} = 24\\\\\frac{x}{4} = 24\\\\x = 24 \times 4\\\\x = 96

Thus there are 96 roses altogether

7 0
2 years ago
Lisa's test grades are 79, 89, and 90. there will be one more test this year. if lisa wants her test average to be at least 88,
erik [133]
For this case, what we are going to do first is to assume that all the exams are worth the same percentage of the final grade.
 We have then that Lisa's average grade point equation is:
 88 =  \frac{79 + 89 + 90 + x}{4}
 Where,
 x: minimum note that lisa must obtain in the last exam.
 Clearing x we have:
 x = (88 * 4) - (79 + 89 + 90)&#10;&#10;x = 94
 Answer:
 
the lowest grade she can get on her last test is:
 
x = 94
5 0
2 years ago
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